Question Number 216786 by mnjuly1970 last updated on 20/Feb/25 Answered by mehdee7396 last updated on 21/Feb/25 $${suppose}\:\:{a},{b},{c}\in{A} \\ $$$${a}+{b}={p}\:\:\left({i}\right)\:\:\&\:\:{a}+{c}={q}\:\:\left({ii}\right)\:\:\&\:\:\:{b}+{c}={m}\:\:\left({iii}\right)\:;\:{p},{q},{m}\in{Q} \\ $$$$\left({i}\right)−\left({ii}\right)\rightarrow{b}−{c}={p}−{q}\:\:\left({iv}\right) \\ $$$$\:\left({iv}\right)+\left({iii}\right)\rightarrow\mathrm{2}{b}={p}−{q}+{m}\in{Q}\:\:\: \\ $$$$\Rightarrow{b}\in{Q}\:\:{it}\:{is}\:\:{incorrect}\:…
Question Number 216734 by mnjuly1970 last updated on 17/Feb/25 Answered by sniper237 last updated on 18/Feb/25 $${Let}\:{E}=\left\{{f}:{A}\rightarrow{B}\right\}\:\:,\:\mid{E}\mid=\mathrm{3}^{{m}} \: \\ $$$${E}\backslash{S}=\left\{{f}\in{E},\:\:\mid{f}\left({A}\right)\mid=\mathrm{1}\:{or}\:\mathrm{2}\right\} \\ $$$$\mid{E}\backslash{S}\mid=\:{C}_{\mathrm{3}} ^{\mathrm{1}} .\mathrm{1}^{{m}} +{C}_{\mathrm{3}}…
Question Number 216739 by ArshadS last updated on 17/Feb/25 $${Prove}\:{that}\:\:^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:−^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\:=\mathrm{1} \\ $$$${Question}#\mathrm{216694}\:{reposted}\:{for}\:{new}\:{answers} \\ $$ Answered by Rasheed.Sindhi last updated on 17/Feb/25 $$\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:=\mathrm{1} \\…
Question Number 216703 by LMKhit last updated on 16/Feb/25 Answered by issac last updated on 16/Feb/25 $$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{h}} }{{h}!\left(\frac{\mathrm{1}}{\mathrm{2}}−{h}\right)!}{x}^{\mathrm{2}{h}} \:,\:\mid{x}\mid\leq\mathrm{1} \\ $$$$\mathrm{because}\:{r}=\underset{{n}\rightarrow\infty} {\mathrm{lim}inf}\mid\:\frac{{a}_{{n}+\mathrm{1}}…
Question Number 216710 by universe last updated on 16/Feb/25 $$\:\:\:\mathrm{let}\:{y}_{\mathrm{1}} \:,\:{y}_{\mathrm{2}} \:,\:{y}_{\mathrm{3}\:\:} …\:{y}_{{p}} \:\mathrm{be}\:\mathrm{fixed}\:\mathrm{positive}\:\mathrm{number} \\ $$$$\:\:\:\mathrm{consider}\:\mathrm{the}\:\mathrm{sequences} \\ $$$$\:{s}_{{n}} \:=\:\:\frac{{y}_{\mathrm{1}} ^{{n}} +{y}_{\mathrm{2}} ^{{n}} +{y}_{\mathrm{3}} ^{{n}} +…+{y}_{{p}}…
Question Number 216661 by MrGaster last updated on 14/Feb/25 $$\mathrm{Prove}:\langle{x}^{{m}} \rangle=\underset{\left\{{p}_{{n}} \right\}} {\acute {\sum}}\overset{{n}} {\prod}\frac{{m}!}{{p}_{{n}} !\left({n}!\right)^{{p}_{{n}} } }\langle{x}^{{n}} \rangle_{{c}} ^{{p}_{{n}} } \\ $$ Terms of…
Question Number 216644 by Raajaravwan last updated on 13/Feb/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216646 by York12 last updated on 13/Feb/25 $$\mathrm{Let}\:{a},{b},{c}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{an}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{such}\:\mathrm{that} \\ $$$${ab}+{bc}+{ac}+{abc}=\mathrm{4},\:\mathrm{show}\:\mathrm{using}\:\mathrm{vieta}'\mathrm{s}\:\mathrm{relations}\:\mathrm{that} \\ $$$$\frac{{a}}{{a}+\mathrm{2}}+\frac{{b}}{{b}+\mathrm{2}}+\frac{{c}}{{c}+\mathrm{2}}=\mathrm{1} \\ $$ Answered by A5T last updated on 14/Feb/25 $$\frac{\mathrm{a}\left(\mathrm{b}+\mathrm{2}\right)\left(\mathrm{c}+\mathrm{2}\right)+\mathrm{b}\left(\mathrm{a}+\mathrm{2}\right)\left(\mathrm{c}+\mathrm{2}\right)+\mathrm{c}\left(\mathrm{a}+\mathrm{2}\right)\left(\mathrm{b}+\mathrm{2}\right)}{\left(\mathrm{a}+\mathrm{2}\right)\left(\mathrm{b}+\mathrm{2}\right)\left(\mathrm{c}+\mathrm{2}\right)}=\mathrm{1} \\…
Question Number 216585 by MathematicalUser2357 last updated on 11/Feb/25 $$\mathrm{is}\:\mathrm{this}\:\mathrm{right}? \\ $$$$\mathrm{i}\:\mathrm{had}\:\mathrm{let}\:\theta=\begin{cases}{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}>\mathrm{0},{d}>\mathrm{0}\right)}\\{\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}<\mathrm{0},{d}>\mathrm{0}\right)}\\{−\pi+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}<\mathrm{0},{d}<\mathrm{0}\right)}\\{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}>\mathrm{0},{d}<\mathrm{0}\right)}\end{cases}\:\mathrm{before}\:\mathrm{i}\:\mathrm{calculated}\:\mathrm{below} \\ $$$$\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{di}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\theta} \\ $$$$=\mid{a}+{bi}\mid^{{c}} \mid{a}+{bi}\mid^{{di}} {e}^{{ic}\theta} {e}^{−{d}\theta}…
Question Number 216582 by klipto last updated on 11/Feb/25 $$ \\ $$$$\mathrm{using}\:\mathrm{first}\:\mathrm{principle}\:\mathrm{solve} \\ $$$$\mathrm{y}=\frac{\mathrm{x}+\mathrm{2}}{\:\sqrt{\mathrm{x}}+\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{with}\:\mathrm{first}\:\mathrm{principle} \\ $$ Commented by Rasheed.Sindhi last updated on 12/Feb/25…